What is the magnifying power of focal length 10cm least distance of distinct vision is 25cm?

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Answer:

According to the question, we have –

Focal length of the given objective lens is f1=2.0 cm

Focal length of the given eyepiece is f2=6.25 cm

Distance between the eyepiece and the objective lens is d =15 cm

(a) Least distance of distinct vision is denoted by d’ = 25 cm

Image distance for the eyepiece,

$v_{2}$= -25cm

Object distance for the eyepiece= is represented by u2

Using the lens formula, we have the relation:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

$\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}$

$\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$

$u_{2}=-5cm$

Now, the image distance for the objective lens becomes –

 $v_{1}=d+u_{2}=15 – 5=10 cm$

Object distance for the objective lens is denoted by  $u_{1}$

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$

$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$

$\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$

$u_{1}=-2.5cm$

Now for the Magnifying power –

$m=\frac{v_{0}}{|u_{0}|}(1+\frac{d}{f_{e}}) $

$=\frac{10}{2.5}(1+\frac{25}{6.25}) $

$=20$

Hence, the magnifying power of the microscope is 20.

(b) The final Image formed is at infinity.

Therefore image distance for the eyepiece is

 $v_{2}=\infty$

Object distance for the eyepiece is denoted by $ u_{2}$

With respect to the lens formula, we have the relation:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

$\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}$

$u_{2}=-6.25cm$

Now, image distance for the objective lens becomes –

 $v_{1} = d + u_{2}=15 -6.25 = 8.75 cm$

Object distance for the objective lens=u1

From the lens formula, we have –

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$

$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$

$\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$

$u_{1}=-\frac{17.5}{6.75}=-2.59cm$

Magnitude of the object distance, $|u_{1}|$=2.59 cm

The magnifying power of a compound microscope is given by the relation –

m=$\frac{v_{1}}{|u_{1}|}(\frac{d’}{|u_{2}|})$

=$\frac{8.75}{2.59}\times\frac{25}{6.25}$

=13.51

Therefore, 13.51 is the magnifying power of the microscope.